14
$\begingroup$

I'm working on a project that involves reading RFID Tags and comparing the signal strength the reader sees when you change the antenna configuration (number of antenna, position, etc...). As part of the project, I need to compare the setups to see which are most effective.

Ideally, I would be able to perform either an Unpaired t-Test or an ANOVA between two antenna positions (or MANOVA between multiple). However, as the response is in decibels which are logarithmic, I'm wondering what the best way to proceed with that is?

Would it be best to convert the results into a linear scale and then compare using one of the methods I mentioned, or should I use decibels as they are with a different statistical test to compare them?

$\endgroup$
3
  • 2
    $\begingroup$ As you are using tests assuming a Gaussian distribution, if the distribution of responses is "more Gaussian" in dB than in the linear scale (i.e. the original data is approximately log normal), it makes sense to stay in the logarithmic scale. $\endgroup$
    – Luca Citi
    Aug 7, 2018 at 19:28
  • 1
    $\begingroup$ @NickCox, I think mathematical-statistics works just fine when requesting a proof, the corresponding tag being a synonym of the former tag. $\endgroup$ Aug 8, 2018 at 17:10
  • 1
    $\begingroup$ Perhaps I should have said "a useless tag for this kind of question". $\endgroup$
    – Nick Cox
    Aug 8, 2018 at 17:12

6 Answers 6

7
$\begingroup$

Whether to transform depends on what scale you want your inference at.

Generally, the variance of a function of $x$ does not equal the function of the variance of $x$. Because $\sigma^{2}_{f(x)} \ne f(\sigma^{2}_{x})$ transforming $x$ with $f$, then performing statistical inference (hypothesis tests or confidence intervals) on $f(x)$, then back-transforming—$f^{-1}$—the results of that inference to apply to $x$ is invalid (since both test statistics and CIs require an estimate of the variance).

Basing CIs on transformed variables + back-transformation produces intervals without the nominal coverage probabilities, so back-transformed confidence about an estimate based on $f(x)$ is not confidence on an estimate based on $x$.

Likewise, inferences about untransformed variables based on hypothesis tests on transformed variables means that any of the following can be true, for example, when making inferences about $x$ based on some grouping variable $y$:

  1. $x$ differs significantly across $y$, but $f(x)$ does not differ significantly across $y$.

  2. $x$ differs significantly across $y$, and $f(x)$ differs significantly across $y$.

  3. $x$ does not differ significantly across $y$, and $f(x)$ does not differ significantly across $y$.

  4. $x$ does not differ significantly across $y$, but $f(x)$ differs significantly across $y$.

In short, knowing whether $f(x)$ differs significantly across groups of $y$ does not tell you whether $x$ differs across $y$.

So the question of whether to transform those dBs is answered by whether you care about dB or exponentiated dB.

$\endgroup$
0
18
$\begingroup$

Strictly we need to see your data to have any chance of giving definitive advice, but it is possible to guess.

As you say, decibels are already on a logarithmic scale. That is likely to mean, for a variety of physical and statistical reasons, that they are quite likely to behave well by being approximately additive, homoscedastic and symmetrically distributed conditional on predictors. But you may be able to give a physical or engineering argument of how the response should vary as you change your design variables.

I know no possible principle or theory that means that you are obliged to exponentiate them before applying a $t$ test or ANOVA. I would expect that to make statistical behaviour worse, not better.

The same kind of reasoning usually applies to other "pre-transformed" logarithmic scales such as pH or the Richter scale.

PS: No idea what RFID tags are.

$\endgroup$
1
  • 5
    $\begingroup$ RFID's tags are Radio Frequency ID tags... those things in your passport, library materials, chipped credit card, etc. that make token-based ID possible wirelessly. $\endgroup$
    – Alexis
    Aug 7, 2018 at 18:02
3
$\begingroup$

Well, the only way to definitively answer this question is to look at some decibel data -- is there a simple distribution (e.g. Gaussian distribution) which is a good model for that? Or is the exponential of the data a better candidate? My guess is that the non-exponentialized data is more nearly Gaussian and therefore to make any ensuing analysis more straightforward, you should use that, but I'll let you be the judge of it.

I take issue with your proposed analysis, which is to apply a significance test to observed data from different experiments (namely different antenna positions). From considering the physics of this, there must be some difference, perhaps minuscule, perhaps substantial. But a priori there is some difference, therefore with a large enough data set, you must reject the null hypothesis of no difference. Thus the effect of a significance test is only to conclude "you have / you don't have a large data set". That doesn't seem very useful.

More useful would be to quantify the difference between different antenna positions, and perhaps also take into account costs and benefits to decide which position is to be selected. Quantified differences are sometimes called "effect size analysis"; a web search for that should turn up some resources. Costs and benefits come under the heading of utility theory and decision theory; again a search will find some resources.

$\endgroup$
1
  • $\begingroup$ the raw measurements may not be gaussian but the residuals after fitting a model, like a polynomial, may be. $\endgroup$ Jun 19, 2022 at 3:07
3
$\begingroup$

The (logarithmic) Decibel scale is useful because the power of a signal can often be described by a (variable) series (or fluid range) of multiplications.

  • E.g. if a 1cm thick wall reduces the signal to $\frac{1}{10}$ power (10 Decibel reduction).
  • then a 2 cm thick wall reduces the signal to $\frac{1}{100}$ power (20 Decibel redution).
  • and a 3cm thick wall reduces the signal to $\frac{1}{1000}$ power (30 Decibel reduction)
  • etc.

More generally, if you make the wall thickness non-discrete then the signal (if you express it in untransformed units) could be expressed by an exponential function $$P[\textrm{mW}] = P_0 \left( \frac{1}{10} \right)^{L[\textrm{cm}]}$$

This is more simple, if you express the logarithm of the signal power, as a linear function (which, if you wish, requires some definition about the absolute scale, in this case 0dB relates to 1 mW)

$$P[\textrm{dB}] = 10 \left(\log(P_0[\textrm{mW}])-L[\textrm{cm}]\right)$$


Whenever you have a process that is multiplicative like:

$$X \propto e^Y $$

with the parameter $Y$ normal distributed: $$Y \sim N(\mu,\sigma^2)$$

then $X$ has a log-normal distribution and $\log(X)$ (or X expressed in any other logarithmic scale, such as the dB scale) has a normal distribution.


I expect that your error term will be multiplicative like that. That is: the signal strength will be a sum of many normal distributed error terms (e.g. amplifier temperature fluctuations, atmospheric conditions, etc.) that occur in the exponent of the expression for the signal strength.

$$y_{i} = e^{x_i+\epsilon_i}$$

$\endgroup$
0
$\begingroup$

An example supporting the use of decibels as-is:

For process capability (AKA accuracy ratio) Cp = the tolerance range (upper limit - lower limit) divided by 6 x Standard Deviation. When there is only one limit Cpk* is the difference between the mean and the limit (commonly called "margin") divided by 3 x StdDev.

If decibel values are converted to their linear equivalents, then the differences are found using division because a decibel value (dB) is a ratio (P1/P2). The difference in the decibel values is calculated with subtraction.

The test:

The Cpk result for A) a test with a positive lower limit (LL) and test values greater than the limit should be exactly the same as B) using a negative upper limit (UL = -LL) and the same test values x -1.

This is true when unconverted decibel test values are used but not when their linear equivalents are used. When the values are converted, the results are similar for test values close to zero but diverge significantly the "larger" they are (farther from zero). In the latter case the Cpk analysis produces false failures for tests with large positive test values and false positives for "large" negative test values.

*I'm not sure this is the correct term associated with a single-limit test but it's easier to reference in the following description.

$\endgroup$
1
  • $\begingroup$ Sample test values for switch isolation are around 40 dB. 'Isolation' is the difference between the input and the output when it's open and is expected to be a loss. With a signal on the input you measure the output with the switch closed, then with it open, then calculate the difference. But a purist might insist that the transfer function is -40 dB. Either is correct. But when the dB values are converted to their linear equivalents, the standard deviation of the two groups is NOT the same because the linear values are 8 orders of magnitude apart. This is why decibels were invented. $\endgroup$
    – GTaylor
    Sep 26, 2023 at 19:06
0
$\begingroup$

More experiments and research indicate that a better method would be to do the statistical calculations (like standard deviation and mean) with the decibel values, then convert them to percentage values for comparison using the following formula.

% = (10^(dBm/10))-1

for the dB/dBc/dBm (power) group. Use /20 for dBV/dBrc values. (And there are others for acoustics, etc.)

This yields a value in the range of -100% to +$\infty$%.

  • 0 dB = 1 = 0%
  • 3 dB = 2 = 100%, -3 dB = 1/2 = -50%
  • 6 dB = 4 = 300%, -6 dB = 1/4 = -75%
  • 10 dB = 10 = 900%, -10 = 1/10 = -90%

The rationale for this is:

  • The linear and decibel scales are almost the same from 1 to 10 (0 dB = 1, 10 dB = 10). The farther from the 0:1 'center point' the more the two diverge. (30 dB = 1,000; -30 dB = 0.001)
  • Decibel values are ratios as are % values. dB is 10(log(P1/P2)). dBm is 10(log(P1/1 mW)). Converting a group of data values that are far removed from the 0:1 'center' but local to each other, to percentage values allows them to be used with/compared to other groups as if they are all in the same -1 to +1 range.
  • It also creates unitless values.
$\endgroup$
1
  • $\begingroup$ I thought dB was $20\log_{10}\left(\tfrac{\text{measured}}{\text{referece}}\right)$? $\endgroup$
    – Alexis
    Nov 28, 2023 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.